Question

differentiate the following functions.
(a) 4pts. $f(x)=\frac{x^{2}sin(x)}{1 + x^{2}}$
(b) 4pts. $f(x)=sin^{2}(3x)sin(4x^{5})$
(c) 4pts. $f(x)=sqrt{1+sqrt{1+sqrt{1 + x}}}$

Explanation:

Step1: Recall quotient - rule for (a)

If $f(x)=\frac{u(x)}{v(x)}$, then $f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^{2}}$. Here $u(x)=x^{2}\sin(x)$ and $v(x)=1 + x^{2}$.

Step2: Differentiate $u(x)$ in (a)

Using product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x^{2}$, $u^\prime=2x$ and $v=\sin(x)$, $v^\prime=\cos(x)$. So $u^\prime(x)=2x\sin(x)+x^{2}\cos(x)$. And $v^\prime(x)=2x$.

Step3: Recall product - rule for (b)

If $f(x)=u(x)v(x)$, then $f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)$. Here $u(x)=\sin^{2}(3x)$ and $v(x)=\sin(4x^{5})$.

Step4: Differentiate $u(x)$ in (b)

Let $t = \sin(3x)$, then $u(x)=t^{2}$. Using chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=2t = 2\sin(3x)$ and $\frac{dt}{dx}=\cos(3x)\cdot3$. So $u^\prime(x)=2\sin(3x)\cos(3x)\cdot3$. And $v^\prime(x)=\cos(4x^{5})\cdot20x^{4}$.

Step5: Recall chain - rule for (c)

If $y = \sqrt{u}$, then $y^\prime=\frac{u^\prime}{2\sqrt{u}}$. Let $u = 1+\sqrt{1+\sqrt{1 + x}}$, then $f^\prime(x)=\frac{u^\prime}{2\sqrt{u}}$. Re - apply chain - rule for $u^\prime$ and continue the process.

Answer:

(a) $f^\prime(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-2x\cdot x^{2}\sin(x)}{(1 + x^{2})^{2}}$
(b) $f^\prime(x)=2\sin(3x)\cos(3x)\cdot3\cdot\sin(4x^{5})+\sin^{2}(3x)\cdot\cos(4x^{5})\cdot20x^{4}$
(c) $f^\prime(x)=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}$